« ΠροηγούμενηΣυνέχεια »
Again, As Radius
Is to the Log. of the Line AC 47 1.672098 So is the Sine of the Angle C58 9.928420
To the Log. Height of the
Tower 40 Yards.
But as I told you before, the Ground must be Level. However, if it be not, I will fhew you,
Prob. 3. How to take the Height of a Tower, &c. when the Ground either rifes or falls.
Fig. 64 AB is the Tower, CB the Hill whereon you are to take the. Height of the Tower Plant your Inftrument in any place of the Hill, as at C, direct the Sights to A, and take the Angle A Cd; which let be 19 d g 30 Minutes. Take alfo the Angle d CB, which is 45, 30'; then meafure the diftance CB 56 Yards, take 19°30′ there remains 70° 30 for the Angle A, then fay,
As Sine 70
is to the diftance CB 56 Yards
So are both the Angles at C, vix.60°.oc!.
To the Height ofthe Tower 54 Yards.
To take this at two Stations, without coming to the Foot of the Tower, is no more then what has been faid before; for you take your Angles at C, and then meafure to E, and there in the like manner as before, take your Angles again, thereby you may find all the Angles, and the Line AE, then fay,
As the Sine of the Angle ABE
To the Log. of the Height of the Tower,
Pob. 4. How to take the Horizontal-Line of a Hill.
Fig. 65. Suppofe KL MN an Hill, whofe Bafe you would know. Plant your Inftrument at K, and caufe a mark to be fet up at L, fo high above the top of the Hill, as the Inftrument flands from the Ground at K, and take the Argle LKN 58 deg. Measure the diftance KL 16 Chains, 80 Lin ks. Then say,
is to the Line KL16 Chains 80Lin. 3.225309 Sois theSineComplement of K.58 9.724210.
to part of the Base KN 8 Cha. 12.949519 90 Lin.
But if you have occafion to measure the whole Hill, plant again your Instrument at L, (or M) and take your Angle NLM, which let be 46 dig. Measure also the diftance L M 21 Chains: Then fay, As Radius
10 000000 Is to the Line LM 21 Chains 1 222219 So is the Sine of the Angle 9850934
MLN 46 To the part of the Base NM 11.179153.
is Chains 12 Lin. Which 15 Chains 12 Lin, added to 8 Chains 90 Lin. makes 24 Chains, 2 Lin. for the whole Base KM.
I mentioned this way, for to make you understand how to take part of a Hill; for many times your Survey may end on the Side of a Hill. But if you find you are to take in the
! whole Hill, you need not take so much pains as the foriner way. But thus: Take, as before, the Angle K 58 deg. Measure KL. Then at L take the Angle KLM 78 deg. SukHract those 2 from 180 deg. Remains 44 for the Angle at M, Then say,
As she Sine of the Angle M.
Prob. 5. How to take the Altitude of an Ob. ject Amding upon a Hill, inacceffible.
Fig. 66. Suppofe NO to be the Object, and you ftanding at P were requi red to find the heighth thereof.
Firft, upon Paper draw a right Line at pleasure, as QT, and make choice of any Point at pleasure, as at P, for the place of your flanding; then with a Quadrant direc ted to the Top of the Object, you find the Degrees cut to be 40°.52', and then direct the Sights to the Bottom of the Object at O, and let the Degreees cut be 22°.25.
Then upon P, protract an ▲ of 40°52′ and draw the Line P w at pleafure: And an of 22°.25′, and draw the Line P c at plea fure.
Secondly, go forwards in a right Line to wards the Object as at R, 2125 feet; and there direct your Sights to the top of the Object at N, and you will find the degrees cut to be 61° 2′, through which draw a line at pleasure, as RS, croffing the line P w in the point M, the top of the Object: From whence a Perpendicular let fall upon the Ground-line PT, as NK; that Line fhall be equal to the Altitude and of the Hill, and the Object together, (and direct the Sights to O, and draw the Line Rt, cutting the Line Pc in the point 0.)
Now by the Interfections of thefe four Lines, Pw, Rs, Pc, Rr, there is conftituted four Triangles, viz. PNK, and RNK, bo h Right Angled at K, and PNR, and RNO Oblique angled: By the refolving of which from the diftance meafured PR, and the feveral Angles obferved, at R and P, you may find the required Altitude.
I. In the Oblique angled Triangle PNR, there is given, the Angle NPR 40°52′, and the Angle NRP 118° 28'; for it is the Complement of the Angle NRK 61° 82, to 180°. And the Side measured PR, 212.5 Foot, And having the Angles at R and P, the Sum of them is 159°. 10', which take from 180, there will remain 20 50', for the Angle PNR. From which Triangle given, the two other Sides PN, and NR may be found by Axiom 2. thus,
As Sine of the 4 PNR 20 50.
So is the Sine of NPR 40° 52
To the Side NR
And fo is the Sine of the Complement of 61.82' 118 25' (or the L NRP) To the Side PN, 515. 66.
In the Right Angled. Triangle RNK there is given, (1) The Hypothenufe RN = 280 Foot (2) The Angle NRK 61o. 3o2 d. Whereby you may fiud NK, by Cafe 1, of Right Angle plain Triangles; thus